We use the fact that \(S_{q,\mathbf {a}}(\mathbf {c})\)  is multiplicative in \(q\) , as proved by Heath-Brown in the discussion following [1]}. This allows us to write
\(\Sigma (x;\mathbf {c}) = \sum _{\begin{array}{c}q_2\leqslant x\\ q_2|(2A)^{\infty }\end{array}}q_2^{-3}S_{q_2,\mathbf {a}}(\mathbf {c})\sum _{\begin{array}{c}q_1 \leqslant x/q_2\\ \gcd (q_1,2A) = 1\end{array}}q_1^{-3}S_{q_1,\mathbf {a}}(\mathbf {c}),\) 
where the notation \(q_2|(2A)^{\infty }\)  means that every prime dividing \(q_2\)  also divides \(2A\) . By [2]}, we have
\(\sum _{\begin{array}{c}q_1 \leqslant x/q_2\\ \gcd (q_1,2A) = 1\end{array}}q_1^{-3}S_{q_1}(\mathbf {c}) = \sum _{\begin{array}{c}q_1 \leqslant x/q_2 \\ \gcd (q_1,2A) = 1\end{array}}\left(\frac{A}{q_1}\right)\frac{\varphi (q_1)}{q_1},\) 
where \(\varphi \)  denotes the Euler totient function. Applying the Burgess bound as in the proof of [2]}, we obtain
\(\sum _{\begin{array}{c}q_1 \leqslant x/q_2\\ \gcd (q_1,2A) = 1\end{array}}q_1^{-3}S_{q_1,\mathbf {a}}(\mathbf {c})\ll \frac{|A|^{3/16+\epsilon }x^{1/2}}{q_2^{1/2}}.\) 
Returning to (REF ), we have
\(\Sigma (x;\mathbf {c})\ll |A|^{3/16+\epsilon }x^{1/2}\sum _{\begin{array}{c}q_2\leqslant x\\ q_2|(2A)^{\infty }\end{array}}\frac{|S_{q_2,\mathbf {a}}(\mathbf {c})|}{q_2^{7/2}}.\) 
Now [2]} states that \(S_{q,\mathbf {a}}(\mathbf {c}) \ll q^3\prod _{i=1}^4 \gcd (q,a_i,c_i)\) . Moreover, there are \(O(x^{\epsilon }A^{\epsilon })\)  choices for \(q_2 \leqslant x\)  with \(q_2|(2A)^{\infty }\) . Therefore
\(\sum _{\begin{array}{c}q_2\leqslant x\\ q_2|(2A)^{\infty }\end{array}}\frac{|S_{q_2,\mathbf {a}}(\mathbf {c})|}{q_2^{7/2}}\ll x^{\epsilon }A^{\epsilon }\sup _{\begin{array}{c}q|(2A)^{\infty }\end{array}}\left(\frac{\prod _{i=1}^4 \gcd (q,a_i,c_i)^{1/2}}{q^{1/2}}\right).\) 
Let \(p\)  be a prime. We define
\(L_p = \nu _{p}\left(\frac{\prod _{i=1}^4 \gcd (q,a_i,c_i)}{q}\right),\quad k_p = \nu _p(q).\) 
For an index \(i=1,\ldots , 4\) , we let \(m_{i,p}\)  denote the \(i\) th smallest element of \(\nu _p(\gcd (a_1,c_1)),\ldots , \nu _p(\gcd (a_4,c_4))\) .
Then
\(L_p = \sum _{i=1}^4 \left(\min (k_p,m_{i,p}) - k_p\right).\) 
It can be checked that the maximum possible value of \(L_p\)  is attained by choosing \(k_p = m_{3,p}\) , and so \(L_p \leqslant m_{1,p}+ m_{2,p}+m_{3,p}\) . We have \(L_p\leqslant 0\)  unless \(p|\Delta _{\mathbf {c}}(\mathbf {a})\) . Hence the supremum in (REF ) is bounded by
\(\prod _{p|\Delta _{\mathbf {c}}(\mathbf {a})}p^{(m_{1,p}+m_{2,p}+m_{3,p})/2}.\) 
For any prime \(p\) , we have
\(\nu _p(\Delta _{\mathbf {c}}(\mathbf {a})) = m_{1,p}+m_{2,p}+m_{3,p}+\min (m_{1,p}+m_{2,p}+m_{3,p},m_{4,p}) \geqslant \frac{4}{3}(m_{2,p}+m_{3,p}),\) 
and hence the supremum in (REF ) is bounded by \(\Delta _{\mathbf {c}}(\mathbf {a})^{1/3}\) . Recalling (REF ) we obtain the desired estimate for \(\Sigma (x;\mathbf {c})\) .
